Consider the situation illustrated in Fig. 5-8(a). The...

Consider the situation illustrated in Fig. $5-8(a)$. The uniform $0.60-\mathrm{kN}$ beam is hinged at $P$. Find the tension in the tie rope and the components of the reaction force exerted by the hinge on the beam. Give your answers to two significant figures. The reaction forces acting on the beam are shown in Fig. $5-8(b)$, where the force exerted by the hinge is represented by its horizontal and vertical components, $F_{R H}$ and $F_{R V}$. The torque equation about $P$ is $$ \text { \&+ } \sum \tau_{P}=+(3 L / 4)\left(F_{T}\right)\left(\sin 40^{\circ}\right)-(L)(800 \mathrm{~N})\left(\sin 90^{\circ}\right)-(L / 2)(600 \mathrm{~N})\left(\sin 90^{\circ}\right)=0 $$ (We take the axis at $P$ because then $F_{R H}$ and $F_{R V}$ do not appear in the torque equation.) Solving this equation yields $F_{T}=2280 \mathrm{~N}$ or, to two significant figures, $F_{T}=2.3 \mathrm{kN}$. To find $F_{R H}$ and $F_{R V}$, write $$ \begin{array}{lll} \pm \sum F_{x}=0 & \text { or } & -F_{T} \cos 40^{\circ}+F_{R H}=0 \\ +\uparrow \sum F_{y}=0 & \text { or } & F_{T} \sin 40^{\circ}+F_{R V}-600-800=0 \end{array} $$ Since we know $F_{T}$, these equations lead to $F_{R H}=1750 \mathrm{~N}$ or $1.8 \mathrm{kN}$ and $F_{R V}=65.6 \mathrm{~N}$ or $66 \mathrm{~N}$.

Consider the situation illustrated in Fig. $5-8(a)$. The uniform $0.60-\mathrm{kN}$ beam is hinged at $P$. Find the tension in the tie rope and the components of the reaction force exerted by the hinge on the beam. Give your answers to two significant figures.
The reaction forces acting on the beam are shown in Fig. $5-8(b)$, where the force exerted by the hinge is represented by its horizontal and vertical components, $F_{R H}$ and $F_{R V}$. The torque equation about $P$ is
$$
\text { \&+ } \sum \tau_{P}=+(3 L / 4)\left(F_{T}\right)\left(\sin 40^{\circ}\right)-(L)(800 \mathrm{~N})\left(\sin 90^{\circ}\right)-(L / 2)(600 \mathrm{~N})\left(\sin 90^{\circ}\right)=0
$$
(We take the axis at $P$ because then $F_{R H}$ and $F_{R V}$ do not appear in the torque equation.) Solving this equation yields $F_{T}=2280 \mathrm{~N}$ or, to two significant figures, $F_{T}=2.3 \mathrm{kN}$.
To find $F_{R H}$ and $F_{R V}$, write
$$
\begin{array}{lll}
\pm \sum F_{x}=0 & \text { or } & -F_{T} \cos 40^{\circ}+F_{R H}=0 \\
+\uparrow \sum F_{y}=0 & \text { or } & F_{T} \sin 40^{\circ}+F_{R V}-600-800=0
\end{array}
$$
Since we know $F_{T}$, these equations lead to $F_{R H}=1750 \mathrm{~N}$ or $1.8 \mathrm{kN}$ and $F_{R V}=65.6 \mathrm{~N}$ or $66 \mathrm{~N}$.

Key Concepts

Force Decomposition

Force decomposition involves breaking down a force vector into its orthogonal (usually horizontal and vertical) components. This process simplifies the analysis of forces acting at angles and is vital for setting up separate equilibrium equations in different directions.

Torque (Moment) and Moment Equilibrium

Torque, or moment of a force, measures the tendency of a force to rotate a body about a pivot or axis and is calculated as the product of the force and the perpendicular distance from the pivot to the line of action of the force. Moment equilibrium requires that the sum of all torques about any point is zero, which is an essential condition for a body in static equilibrium.

Reaction Forces at Supports

Reaction forces are the forces developed at supports or connections (such as hinges) to balance the external loads acting on a structure. Understanding how to determine the horizontal and vertical components of these reaction forces is critical in ensuring that the structure satisfies the conditions for static equilibrium.

Static Equilibrium

Static equilibrium refers to the state of a body where it is at rest or moves with constant velocity, meaning that the net force and the net moment (torque) acting on it are zero. This principle is essential in analyzing structures because it allows us to set up equations based on the balance of forces and moments, enabling the determination of unknown forces or reactions acting on the body.

Free Body Diagram

A free body diagram is a simplified representation of a body isolated from its environment, showcasing all external forces and moments acting on it. This tool is crucial in mechanics as it helps in visualizing and setting up the equilibrium equations needed to solve for unknown quantities, such as reaction forces, tensions, or other applied loads.

Transcript

00:01 Here in this given problem first of all this is the wall, then the hinged beam, uniform beam fixed horizontally and then the cable, the rope attached with the wall and with the beam.

00:23 If the length of this beam is supposed to be l as it is uniform, so its weight will be acting at its midpoint at a distance of l by 2 from the hinge and this weight is given as 0 .60 kilo newton or we can say this is 600 newton.

00:50 Then the weight attached at its end point that is 800 newton and this point at which the rope is attached that is 3l by 4 from the hinge.

01:07 This is the tension force in the rope.

01:11 Suppose the two components of the tension force in the hinge, horizontal force fx and vertical component fy, angle made by the rope with the beam that is given as 40 degree.

01:30 So, the two components of this tension force, vertical component t sin 40 degree and horizontal component t cos 40 degree.

01:47 We have to find tension force t and fx and fy.

01:53 So, first of all taking equilibrium of rotation of the beam about the hinge, this will be the counter clockwise torque created by t sin 40 degree and torque will be given by the product of force with the perpendicular distance 3l by 4 from the hinge is equal to clockwise torques, two clockwise torques.

02:47 One created by weight of the beam 600 multiplied by l by 2 plus another created by the weight suspended with the beam 800 multiplied by l.

03:02 Here this length will be cancelled from both the sides and rearranging the terms we get 3t sin 40 degree divided by 4 is equal to 1100.

03:19 So, this t is calculated to get its value to be equal to 2281 .7 newton or we can say this tension force in the rope that is 2 .3 kilo newton, one of the answer for this given problem here...

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